Answer:
3/2 and -4
Step-by-step explanation:
that is the answer
Item 10 Find the median, first quartile, third quartile, and interquartile range of the data. 38,55,61,56,46,67,59,75,65,58
jok3333 [9.3K]
I think the median is 58.5, the first quartile is 50.5, the third quartile is 66, and finally, the interquartile range is <span>12.75.</span>
You can solve this problem through factoring.
First, you have the equation,
![h(x) = \frac{x^2-36}{x-6}](https://tex.z-dn.net/?f=h%28x%29%20%3D%20%5Cfrac%7Bx%5E2-36%7D%7Bx-6%7D)
Then, you can factor the numerator.
![h(x) = \frac{(x+6)(x-6)}{x-6}](https://tex.z-dn.net/?f=h%28x%29%20%3D%20%5Cfrac%7B%28x%2B6%29%28x-6%29%7D%7Bx-6%7D)
You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.
You are left with ![h(x) = x+6](https://tex.z-dn.net/?f=h%28x%29%20%3D%20x%2B6)
The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.
To find the number or rows, divide the total number of students by the number in each row.
63 / 7 = 9 rows.
Answer:
Maybe try doing it earlier, don’t try doing things last minute it will put too much stress. Thank you :)
Step-by-step explanation: